If fx=xα log x, x>0 0 , x=0and Rolle's theorem is applicable to fx x∈0. 1 then α is equal to

∴RHL=limx→0+ f(x)=0 LHL=limx→0+ log⁡xx−α(L' Hospital rule) =limx→0+ 1x−αx−α−1=limx→0+ xα−α For the limit to exist α>0 For α>0.limx→0+ f(x)=0 α=12f(x) is continuous for ⇒α>0 α=12

If fx=xα log x, x>0 0 , x=0and Rolle's theorem is applicable to fx x∈0. 1 then α is equal to

∴RHL=limx→0+ f(x)=0 LHL=limx→0+ log⁡xx−α(L' Hospital rule) =limx→0+ 1x−αx−α−1=limx→0+ xα−α For the limit to exist α>0 For α>0.limx→0+ f(x)=0 α=12f(x) is continuous for ⇒α>0 α=12

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$If f (x) = \{\begin{cases} x^{α} \log x, x > 0 \\ 0, x = 0 \end{cases} and Rolle' s theorem is applicable to f (x) x \in [0 . 1] then α is equal to$

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Detailed Solution

$∴ R H L = lim_{x \to 0^{+}} f (x) = 0 L H L = lim_{x \to 0^{+}} \frac{\log x}{x^{- α}} (L' H o s p i t a l r u l e) \begin{aligned} = lim_{x \to 0^{+}} \frac{\frac{1}{x}}{- α x^{- α - 1}} \\ = lim_{x \to 0^{+}} \frac{x^{α}}{- α} F o r t h e l i m i t t o e x i s t α > 0 For α > 0 . lim_{x \to 0^{+}} f (x) = 0 α = \frac{1}{2} \end{aligned}$
f(x) is continuous for
$\Rightarrow α > 0 α = \frac{1}{2}$